Ta dùng công thức \(1+2+...+n=\dfrac{n\times\left(n+1\right)}{2}\). Khi đó
\(\dfrac{1}{1+2}=\dfrac{1}{\dfrac{2\times3}{2}}=\dfrac{2}{2\times3}\);
\(\dfrac{1}{1+2+3}=\dfrac{1}{\dfrac{3\times4}{2}}=\dfrac{2}{3\times4}\);
\(\dfrac{1}{1+2+3+4}=\dfrac{1}{\dfrac{4\times5}{2}}=\dfrac{2}{4\times5}\);
...;
\(\dfrac{1}{1+2+3+...+2020}=\dfrac{1}{\dfrac{2020\times2021}{2}}=\dfrac{2}{2020\times2021}\).
\(\Rightarrow\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+2020}\)
\(=\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+\dfrac{2}{4\times5}+...+\dfrac{2}{2020\times2021}\)
\(=2\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{2020\times2021}\right)\)
\(=2\left(\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+...+\dfrac{2021-2020}{2020\times2021}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2021}\right)\)
\(=\dfrac{2019}{2021}\)
